Python numpy array manipulation

Question

i need to manipulate an numpy array:

My Array has the followng format:

x = [1280][720][4]

The array stores image data in the third dimension:

x[0][0] = [Red,Green,Blue,Alpha]

Now i need to manipulate my array to the following form:

x = [1280][720]
x[0][0] = Red + Green + Blue / 3

My current code is extremly slow and i want to use the numpy array manipulation to speed it up:

for a in range(0,719):
    for b in range(0,1279):
        newx[a][b] = x[a][b][0]+x[a][b][1]+x[a][b][2]
x = newx

Also, if possible i need the code to work for variable array sizes.

Thansk Alot

bastelflp · Accepted Answer · 2016-02-06 17:39:00Z

1

Use the numpy.mean function:

import numpy as np

n = 1280
m = 720

# Generate a n * m * 4 matrix with random values 
x = np.round(np.random.rand(n, m, 4)*10)

# Calculate the mean value over the first 3 values along the 2nd axix (starting from 0) 
xnew = np.mean(x[:, :, 0:3], axis=2)

x[:, :, 0:3] gives you the first 3 values in the 3rd dimension, see: numpy indexing
axis=2 specifies, along which axis of the matrix the mean value is calculated.

answered Feb 6, 2016 at 17:39

bastelflp

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4 Comments

Tim Heß Over a year ago

Thanks for the help, is there a way to keep my 3 dimensions. So xnew[0][0]=[x] ?

bastelflp Over a year ago

Use: xnew_3 = x[:, :, 0:3]

Tim Heß Over a year ago

Sorry if I was unclear, I ment to store the mean value in the third dimension. x[1280,720,1] @bastelflp

bastelflp Over a year ago

Resize the first array and overwrite the 3rd dimension with the mean values: xnew_3 = np.resize(x, (x.shape[0], x.shape[1], 1)) and xnew_3[:, :, 0] = xnew

Brian Huey · Accepted Answer · 2016-02-06 17:40:06Z

0

Slice the alpha channel out of the array, and then sum the array along the RGB axis and divide by 3:

x = x[:,:,:-1]
x_sum = x.sum(axis=2)
x_div = x_sum / float(3)

answered Feb 6, 2016 at 17:40

Brian Huey

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